Question: $f(x) = \begin{cases} 4-4x & \text{for} ~~~~x\gt{1} \\ 6x^2-6& \text{for} ~~~~ x \leq1\end{cases}$ Evaluate the definite integral. $\int^4_{0}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $-22$ (Choice B) B $-16$ (Choice C) C $14$ (Choice D) D $20$
Explanation: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^4_{0}f(x)\,dx$ $= \int^4_{1}f(x)\,dx + \int^{1}_{0}f(x)\,dx~~~~~~$ [Why did we split at 1?] $= \int^4_{1}(4-4x)\,dx + \int^{1}_{0}(6x^2-6)\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^4_{1}(4-4x)\,dx &=4x-2x^2\Bigg|^4_{{1}} \\\\ &= \left[4\cdot(4)-2\cdot ( 4)^2 \right] - \left[4\cdot(1)-2\cdot ( 1)^2\right] \\\\ &= \left[-16\right] -\left[2 \right] \\\\ &= {-18}\end{aligned}$ The second definite integral: $\begin{aligned} \int^{1}_{0}(6x^2-6)\,dx &=(2x^3-6x)\Bigg|^1_{{0}} \\\\ &= \left[ 2\cdot( 1)^3 -6\cdot(1)\right] - \left[2\cdot( 0)^3 -6\cdot(0)\right] \\\\ &= \left[-4\right] -\left[0 \right] \\\\ &= {-4}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^4_{1}(4-4x)\,dx + \int^{1}_{0}(6x^2-6)\,dx$ $ = {-18} + ({-4})$ $ = -22$ The answer $\int^4_{0}f(x)\,dx = -22$